Part Two - 4x ^{3}Part Three - 27x ^{4}Part Four - 108x ^{5}Back to Equilibrium Menu

The general problem is this:

Given the K_{sp}, calculate the molar solubility (in mol/L) of a saturated solution of the substance.

**Example #1:** Silver chloride, AgCl, has a K_{sp} = 1.77 x 10¯^{10}. Calculate its solubility in moles per liter.

**Solution:**

1) Write the dissociation equation:

AgCl(s) ⇌ Ag^{+}(aq) + Cl¯(aq)

2) Write the K_{sp} expression:

K_{sp}= [Ag^{+}] [Cl¯]

This is the equation we must solve.

3) Put in the K_{sp} value:

1.77 x 10¯^{10}= [Ag^{+}] [Cl¯]

4) We have to reason out the values of the two guys on the right. I do that by first assigning a variable, s, to the molar solubility of AgCl. That's the value that I want to determine.

To do this, I look at the molar relationship between AgCl and Ag^{+}. (I do this because I want to express the concentrations of the ions by using 's.') I note that the molar relationship is a 1:1 molar ratio, meaning that for every one mole of AgCl that dissolves, one mole of Ag^{+}is produced. That leads me to this:[Ag^{+}] = sNow, I look at the relationship between AgCl and Cl¯. I see that it is also a 1:1 molar ratio, leading me to this:

[Cl¯] = sI am now ready to substitute into the K

_{sp}expression.

5) Substituting, we get:

1.77 x 10¯^{10}= (s) (s)

7) Now, we take the square root of __both sides__. I hope I'm not too insulting when I emphasize both sides. I have had lots of people in my classes take the square root of the x^{2} side, but not the other. After the square root, we get:

s = 1.33 x 10¯^{5}M

This is the answer because there is a one-to-one relationship between the Ag^{+} dissolved and the AgCl it came from. So, the molar solubility of AgCl is 1.33 x 10¯^{5} moles per liter.

One last thing. The K_{sp} value does not have any units on it, but when you get to the value for s, be sure to put M (for molarity) on it. The reasons behind this are complex and beyond the scope of the ChemTeam's goals for this web site.

Comment: it's important that you know how substances ionize. AgCl is easy, but you may not know that CuSCN (example #4) and silver azide (example #5) also ionize in a 1:1 molar ratio like AgCl.

Warning: You may know lots of common ions but, in a problem like the one under discussion, you may get an unusual one thrown at you on the test. Be prepared!

**Example #2:** Aluminum phosphate has a K_{sp} of 9.83 x 10¯^{21}. What is its molar solubility in pure water?

**Solution:**

1) Here is the dissociation equation:

AlPO_{4}(s) ⇌ Al^{3+}(aq) + PO_{4}^{3}¯(aq)

2) Here is the K_{sp} expression:

K_{sp}= [Al^{3+}] [PO_{4}^{3}¯]

3) Keep in mind that the key point is the one-to-one ratio of the ions in solution. This means that the two ions are equal in their concentration. That allows this equation:

9.83 x 10¯^{21}= (s) (s)

4) Solving gives:

s = 9.91 x 10¯^{11}M

which is the answer.

**Example #3:** Calculate the molar solubility of barium sulfate, K_{sp} = 1.07 x 10¯^{10}

**Solution:**

1) The dissociation equation and the K_{sp} expression:

BaSO_{4}(s) ⇌ Ba^{2+}(aq) + SO_{4}^{2}¯(aq)

K_{sp}= [Ba^{2+}] [SO_{4}^{2}¯]

2) The equation:

1.07 x 10¯^{10}= (s) (s)

3) when solved, gives:

s = 1.03 x 10¯^{5}M

The answer.

Remember, this is the answer because the dissolved ions and the solid are also in a one-to-one molar ratio.

Notice how I did not say 'saturated solution' in the problem. When you see this, you need to assume that it is a saturated solution. Anything else makes the problem unworkable and that is not the intent of the question writer.

**Example #4:** CuSCN, K_{sp} = 1.77 x 10¯^{13}

CuSCN(s) ⇌ Cu^{+}(aq) + SCN¯(aq)molar solubility = 4.21 x 10¯

^{7}M

**Example #5:** Silver azide has the formula AgN_{3} and K_{sp} = 2.0 x 10¯^{8}

AgN_{3}(s) ⇌ Ag^{+}(aq) + N_{3}¯(aq)Note azide, a fairly uncommon polyatomic ion. It could show up on the test! Calculating the molar solubility is left to the student.

**Bonus Example:** Magnesium ammonium phosphate, MgNH_{4}PO_{4}, Ksp = 2.5 x 10¯^{13}

This is how the substance ionizes:

MgNH_{4}PO_{4}(s) ⇌ Mg^{2+}(aq) + NH_{4}PO_{4}^{2}¯(aq)

The ammonium phosphate polyatomic ion, NH_{4}PO_{4}^{2}¯, is an uncommon one.

Part Two - 4x ^{3}Part Three - 27x ^{4}Part Four - 108x ^{5}Back to Equilibrium Menu